Let $$\vec{c}$$ be a vector perpendicular to the vectors $$\vec{a} = \hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$$. If $$\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$$, then the value of $$\vec{c} \cdot (\vec{a} \times \vec{b})$$ is equal to ________.

We need $$\vec{c}$$ perpendicular to both $$\vec{a} = \hat{i} + \hat{j} - \hat{k}$$ and $$\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$$. So $$\vec{c}$$ is parallel to $$\vec{a} \times \vec{b}$$.

Computing $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(1+2) - \hat{j}(1+1) + \hat{k}(2-1) = 3\hat{i} - 2\hat{j} + \hat{k}$$.

So $$\vec{c} = \lambda(3\hat{i} - 2\hat{j} + \hat{k})$$ for some scalar $$\lambda$$.

Using $$\vec{c} \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 8$$: $$\lambda(3 - 2 + 3) = 8$$, so $$4\lambda = 8$$, giving $$\lambda = 2$$.

Therefore $$\vec{c} = 6\hat{i} - 4\hat{j} + 2\hat{k}$$.

Now $$\vec{c} \cdot (\vec{a} \times \vec{b}) = (6\hat{i} - 4\hat{j} + 2\hat{k}) \cdot (3\hat{i} - 2\hat{j} + \hat{k}) = 18 + 8 + 2 = 28$$.

The answer is $$28$$.